To establish the equivalence of
our definitions and the conventional thermodynamic ones we shall make contact with the zeroth law of thermodynamics.
This law has
an analogy with mechanics, where in equilibrium the forces are balanced.
In particular if two subsystems (which we label 1 and 2) are in contact and
in equilibrium:
T1
=
P1
=
(7)
=
The zeroth law has a
fairly straightforward
statistical interpretation and this will allow us to begin to establish the equivalence
between the statistical definitions and the conventional thermodynamic ones.
Consider two systems that are free to exchange energy but
are isolated from the rest of the universe by an ideal
insulating surface. The particle numbers N1, N2 and volumes
V1, V2 are fixed for each subsystem. The total energy will be
constant under our assumptions and we assume further that the two
subsystems are sufficiently weakly interacting that
E = E1 + E2,
(8)
where E1 and E2 are the energies of the subsystems. Assume that the
densities of state
g(E), g1(E),g2(E) are coarse grained so that
,
,
.
We then have
(9)
If the subsystems are sufficiently large, the product
g2(E-E1)g1(E1) will be a sharply peaked function of
E1. From the definition of the entropy we note that it is a monotonically increasing function of g and that
the product g1g2 will be at a maximum when the total
entropy
S(E, E1) = S1(E1) + S2(E-E1)
(10)
is at a maximum. The most likely value
of E1
is the one for which
(11)
Since
,
we find using (), that
(12)
or T1 = T2. The most probable partition of energy between the
two systems is the one for which the two temperatures are the same.
Consider next two subsystems that are separated by a movable wall.
The two
systems are free to exchange energy, but the number of particles
is fixed in each subsystem and the total volume V=V1+V2 is
constant. We write
E=E1+E2. The density of allowed states for
the total system is then
The integrand is sharply peaked for large systems and takes on its
maximum value when
S1(E1,V1)+S2(E-E1,V-V1)=max. Differentiation
using
implies that it is overwhelmingly probable that the system will
be near a state for which
or
T1=T2,P1=P2. Conventionally, one would say that the
pressure in the two compartments must be equal at equilibrium
because the forces have to be in balance. The argument now being
made is quite different, there are no forces, instead the movable
wall is guided to its equilibrium by the invisible hand of the law of
large numbers.
Similarly, consider two systems 1 and 2 which are free to exchange
particles and energy. It is easy to show that the most probable
configuration is the one for which
Next:Boltzmann factor Up:Boltzmann statistics Previous:Statistical definition of thermodynamicBirger Bergersen 1998-09-14